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This question already has an answer here:

You have two containers with equal volume of mutually soluble liquids A and B. You take a table spoon of A and pour it into B, mix, and then the same table spoon of B back into A.

Now which one has a higher concentration of the other liquid? Why? (assume no loss of volume when mixing)

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marked as duplicate by Gareth McCaughan, JonMark Perry, Deusovi, Gamow, Engineer Toast Jun 24 '16 at 12:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Initially, the situation is like this:

                  ┌───────────────┬────────────────┬───────────────┐
                  │  Container 1  │      Spoon     │  Container 2  │
┌─────────────────┼───────────────┼────────────────┼───────────────┤
│ Total volume    │       V       │        0       │       V       │
│ Mass A          │       Ma      │        0       │       0       │
│ Concentration A │      Ma/V     │        -       │       0       │
│ Mass B          │       0       │        0       │       Mb      │
│ Concentration B │       0       │        -       │      Mb/V     │
└─────────────────┴───────────────┴────────────────┴───────────────┘

Taking 1 spoon of volume Vs from container 1

                  ┌───────────────┬────────────────┬───────────────┐
                  │  Container 1  │      Spoon     │  Container 2  │
┌─────────────────┼───────────────┼────────────────┼───────────────┤
│ Total volume    │      V-Vs     │        Vs      │       V       │
│ Mass A          │   Ma(V-Vs)/V  │     Ma*Vs/V    │       0       │
│ Concentration A │      Ma/V     │       Ma/V     │       0       │
│ Mass B          │       0       │        0       │       Mb      │
│ Concentration B │       0       │        0       │      Mb/V     │
└─────────────────┴───────────────┴────────────────┴───────────────┘

Pouring the spoon to container 2, assuming additive volumes,

                  ┌───────────────┬────────────────┬───────────────┐
                  │  Container 1  │      Spoon     │  Container 2  │
┌─────────────────┼───────────────┼────────────────┼───────────────┤
│ Total volume    │      V-Vs     │        0       │      V+Vs     │
│ Mass A          │   Ma(V-Vs)/V  │        0       │    Ma*Vs/V    │
│ Concentration A │      Ma/V     │        -       │ Ma*Vs/V(V+Vs) │
│ Mass B          │       0       │        0       │       Mb      │
│ Concentration B │       0       │        -       │   Mb/(V+Vs)   │
└─────────────────┴───────────────┴────────────────┴───────────────┘

Taking 1 spoon of volume Vs from container 2

                  ┌───────────────┬────────────────┬───────────────┐
                  │  Container 1  │      Spoon     │  Container 2  │
┌─────────────────┼───────────────┼────────────────┼───────────────┤
│ Total volume    │      V-Vs     │       Vs       │       V       │
│ Mass A          │   Ma(V-Vs)/V  │ Ma*Vs²/V(V+Vs) │  Ma*Vs/(V+Vs) │
│ Concentration A │      Ma/V     │  Ma*Vs/V(V+Vs) │ Ma*Vs/V(V+Vs) │
│ Mass B          │       0       │  Mb*Vs/(V+Vs)  │  Mb*V/(V+Vs)  │
│ Concentration B │       0       │    Mb/(V+Vs)   │   Mb/(V+Vs)   │
└─────────────────┴───────────────┴────────────────┴───────────────┘

Pouring the spoon to container 1, again with additive volumes,

                  ┌───────────────┬────────────────┬───────────────┐
                  │  Container 1  │      Spoon     │  Container 2  │
┌─────────────────┼───────────────┼────────────────┼───────────────┤
│ Total volume    │       V       │        0       │       V       │
│ Mass A          │  Ma*V/(V+Vs)  │        0       │  Ma*Vs/(V+Vs) │
│ Concentration A │   Ma/(V+Vs)   │        -       │ Ma*Vs/V(V+Vs) │
│ Mass B          │  Mb*Vs/(V+Vs) │        0       │  Mb*V/(V+Vs)  │
│ Concentration B │ Mb*Vs/V(V+Vs) │        -       │   Mb/(V+Vs)   │
└─────────────────┴───────────────┴────────────────┴───────────────┘

This time the math was a bit more difficult to do mentally. This is the mass of liquid A in container 1:

$M_a \frac{V-V_s}{V} + M_a \frac{V_s^2}{V(V+V_s)} = M_a \frac{V^2-V_s^2}{V(V+V_s)} + M_a \frac{V_s^2}{V(V+V_s)} = M_a \frac{V^2}{V(V+V_s)} = M_a \frac{V}{V+V_s} $

Therefore, the answer is

Container 1 ends up having $M_b \frac{Vs}{V(V+V_c)}$ concentration of liquid B.
Container 2 ends up having $M_a \frac{Vs}{V(V+V_c)}$ concentration of liquid A.
So it depends.

  • If liquid A is more dense than liquid B, i.e. $M_a > M_b$,

    Then container 1 has lower concentration of B than container 2 has of A

  • If liquid A is as dense as liquid B, i.e. $M_a = M_b$,

    Then container 1 has the same concentration of B as container 2 has of A

  • If liquid A is less dense than liquid B, i.e. $M_a < M_b$,

    Then container 1 has higher concentration of B than container 2 has of A

Note I assumed mass concentration, but the result would be the same for molarity. They are the typical measures of concentration.

I think other answers say the concentrations are the same because they use strange measures like percentage of volumes or something like that.

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You can figure this out without using any math at all.

The volume of the containers start as equal. We move one table spoon one way, then the other, so the volumes are still even.
For any amount of liquid A in container B, there must have be moved an equal amount of liquid B to container A for this to be true.

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They will each have the same concentration of the other. Let the amounts of A and B in the second spoonful be x and y, the original volume be V, and one tablespoon be T.
y = T-x, so the amounts in the first jar after the final mixing will be
A: V-T+x (start with V, take away T, and then add x)
B: T-x
Meanwhile, the amount of A that remains in the mostly B jar is the remainder of the tablespoon, I.e. T-x, and the amount of B remaining in the jar is V-(T-x) = V-T+x
Thus, the ratios are the same.

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They are the same
If you take the volume of each as V, concentrations of A and B and a table spoon as t

After you move your A tbsp
B has a concentration of (At)/(V+t).
This concentration won't change any more

You move B back to A and
the concentration of A becomes (B'*t)/(V-t+t)
where B' = (B
V)/(V+t) which is the new concentration of B
Then simplfy the expression and it becomes
B*t/(V+t)

which is the same as the A concentration in B.

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No initial volume is given, so, let's assume that it doesn't matter and pick something convenient, say 2 tablespoons each.

When we add half of A to B, we get a 1/3:2/3 solution. One tablespoon of this contains 1/3 T of A, and 2/3 T of B. Adding that to the 1 T of A that's left, makes the A bottle 2A:1B, while the original B container is 1A:2B.

So, the amount of B that's in A is the same as the amount of A that's in B.

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  • 1
    $\begingroup$ You've only proven it for one scenario, not all scenarios. $\endgroup$ – Trenin Jun 24 '16 at 12:33

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